Precalculus Examples

$2 \cos^{2} (x) - \cos (x) - 3 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = \cos (x)$ . Substitute $u$ for all occurrences of $\cos (x)$ .

$2 u^{2} - u - 3 = 0$

Step 1.2

Factor by grouping.

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Step 1.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 1$ .

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Step 1.2.1.1

Factor $- 1$ out of $- u$ .

$2 u^{2} - (u) - 3 = 0$

Step 1.2.1.2

Rewrite $- 1$ as $2$ plus $- 3$

$2 u^{2} + (2 - 3) u - 3 = 0$

Step 1.2.1.3

Apply the distributive property.

$2 u^{2} + 2 u - 3 u - 3 = 0$

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

$(2 u^{2} + 2 u) - 3 u - 3 = 0$

Step 1.2.2.2

Factor out the greatest common factor (GCF) from each group.

$2 u (u + 1) - 3 (u + 1) = 0$

Step 1.2.3

Factor the polynomial by factoring out the greatest common factor, $u + 1$ .

$(u + 1) (2 u - 3) = 0$

Step 1.3

Replace all occurrences of $u$ with $\cos (x)$ .

$(\cos (x) + 1) (2 \cos (x) - 3) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) + 1 = 0$

$2 \cos (x) - 3 = 0$

Step 3

Set $\cos (x) + 1$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\cos (x) + 1$ equal to $0$ .

$\cos (x) + 1 = 0$

Step 3.2

Solve $\cos (x) + 1 = 0$ for $x$ .

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Step 3.2.1

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 3.2.2

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$x = \arccos (- 1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

The exact value of $\arccos (- 1)$ is $π$ .

$x = π$

Step 3.2.4

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 3.2.5

Subtract $π$ from $2 π$ .

$x = π$

Step 3.2.6

Find the period of $\cos (x)$ .

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Step 3.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = π + 2 π n$ , for any integer $n$

Step 4

Set $2 \cos (x) - 3$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $2 \cos (x) - 3$ equal to $0$ .

$2 \cos (x) - 3 = 0$

Step 4.2

Solve $2 \cos (x) - 3 = 0$ for $x$ .

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Step 4.2.1

Add $3$ to both sides of the equation.

$2 \cos (x) = 3$

Step 4.2.2

Divide each term in $2 \cos (x) = 3$ by $2$ and simplify.

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Step 4.2.2.1

Divide each term in $2 \cos (x) = 3$ by $2$ .

$\frac{2 \cos (x)}{2} = \frac{3}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{2 \cos (x)}{2} = \frac{3}{2}$

Step 4.2.2.2.1.2

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{3}{2}$

Step 4.2.3

The range of cosine is $- 1 \leq y \leq 1$ . Since $\frac{3}{2}$ does not fall in this range, there is no solution.

No solution

Step 5

The final solution is all the values that make $(\cos (x) + 1) (2 \cos (x) - 3) = 0$ true.

$x = π + 2 π n$ , for any integer $n$